Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrating factor for first order linear equations uniqueness theorem

  1. Sep 12, 2011 #1
    My book stated the following theorem: If the functions P(x) and Q(x) are continuous on the open interval I containing the point x0, then the initial value problem dy/dx + P(x)y = Q(x), y(x0)=y0 has a unique solution y(x) on I, given by the formula y=1/I(x)[itex]\int[/itex]I(x)Q(x)dx where I(x) is the integrating factor.

    Now the book showed how the Integrating Factor Method was developed, but it doesn't prove this theorem, particularly why a unique solution exists and why there are no other solutions of a different form (singular solutions).

    Also it states, "The appropriate value of the constant C can be selected "automatically" by writing, I(x)=exp([itex]\int[/itex][itex]^{x_{0}}_{x}[/itex]P(t)dt) and y(x)=1/I(x)[y0+[itex]\int[/itex][itex]^{x_{0}}_{x}[/itex]I(t)Q(t)dt]"

    I don't understand how they got this form. If you have a definite integral there shouldn't be constant like y0 either...
     
  2. jcsd
  3. Sep 12, 2011 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    As far as the proof of existence an uniqueness goes, it is straightforward and can be found at: http://arapaho.nsuok.edu/~okar-maa/news/okarproceedings/OKAR-2006/finan.pdf [Broken]

    As for the definite form of the integrating factor, you are indeed correct, no constant arises from the definite integral and therefore we must add an appropriate constant so that the solution satisfies the boundary conditions.

    We have

    [tex]y(x) = \frac{1}{I(x)}\left(y_0 + \int_x^{x_0} I(t)Q(t)\;\text{d}t\right)[/tex]

    [tex]I(x) = \exp\left(\int_x^{x_0} P(t)\;\text{d}t\right)[/tex]

    So, if we consider [itex]x=x_0[/itex] we find


    [tex]y(x_0) = \frac{1}{I(x_0)}\left(y_0 + \int_{x_0}^{x_0} I(t)Q(t)\;\text{d}t\right)[/tex]

    Obviously the integral vanishes, leaving

    [tex]y(x_0) = \frac{y_0}{I(x_0)}[/tex]

    with

    [tex]I(x_0) = \exp\left(\int_{x_0}^{x_0} P(t)\;\text{d}t\right) = \exp(0) = 1[/tex]

    Hence

    [tex]y(x_0) = y_0[/tex]

    as required.
     
    Last edited by a moderator: May 5, 2017
  4. Sep 12, 2011 #3
    Thanks, that makes sense. One thing I don't understand in the link you gave me is on page four, they state w(t)=y1(t0)-y2(t0)=y0-y0. Why are the intial values of those two functions the same, if we assumed that the two functions would be different?
     
  5. Sep 12, 2011 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Because even though they may be different functions, if they are solutions of the same boundary value problem, then they must satisfy the same boundary conditions!

    Does that make sense?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integrating factor for first order linear equations uniqueness theorem
Loading...